Schur's inequality

Issai Schur (1875 - 1941) was a Jewish mathematician, born in what is now Belarus who studied and worked most of his life in Germany. He died in Tel-Aviv, Israel, two years after emigrating from Germany.

Among many significant results that bear his name, there is a surprising inequality with an instructive one-line proof:

For non-negative real numbers $x,$ $y,$ $z$ and a positive number $t,$

$x^t(x-y)(x-z)+y^t(y-z)(y-x)+z^t(z-x)(z-y)\ge 0.$

The equality holds in two cases:

$x=y=z,$ or

one of them is $0,$ while the other two are equal.

Proof

Because of the symmetry of the left-hand side in the variables $x,$ $y,$ $,$ we may assume without loss of generality that $x\ge y\ge z.$ Rewrite the inequality as

$(x-y)[x^t(x-z)-y^t(y-z)]+z^t(z-x)(z-y)\ge 0$

because, under the assumption $x\ge y\ge z,$ $x^t\ge y^t$ and $x-z\ge y-z,$ such that the two summands on the left are both non-negative.

That the equality holds in the two abovementioned cases is obvious. That these are the only two cases when this happens is more involved.

Generalization

The only place where the positiveness of $t$ has been used was in establishing the implication $x\ge y\Rightarrow x^t\ge y^t.$ But this is true for any monotone increasing function $f$ such that there is an immediate generalization:

$f(x)(x-y)(x-z)+f(y)(y-z)(y-x)+f(z)(z-x)(z-y)\ge 0.$

This is so natural and straightforward that the appearance of the exponents (say, taking $f(u)=e^u)\;$ is sometimes described as a red herring - an artificial distraction from the essence of the problem.